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Sphinxa [80]
2 years ago
11

A chemist adds 255 .0 mL of a M copper(II) sulfate solution to a reaction flask. Calculate the mass in grams of copper(II) sulfa

te the chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
melomori [17]2 years ago
7 0

Answer:

The correct answer is 40.7 grams.

Explanation:

Based on the given information, the volume of copper sulfate added to the solution is 255 ml or 0.255 L. The molarity of copper sulfate is not given, so let us consider it to be 1 M, which can also be written as 1 moles per liter.

The moles of copper sulfate can be determined by using the formula,

Moles = Molarity * Volume in Liters

Moles of CuSO4 = 1 moles/Liter * 0.255 L

Moles of CuSO4 = 0.255 moles

The mass of CuSO4 added in the solution will be,

Mass of CuSO4 = Moles * Molecular mass

= 0.255 moles * 159.609 grams per mole

= 40.7 grams.

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4 0
3 years ago
The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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3 years ago
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Sati [7]
C.) Newton. & it's S.I. Unit of Force.

Hope this helps!
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3 years ago
Read 2 more answers
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
Natali5045456 [20]
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

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3 years ago
Oxidation of Nitrogen in NO2​
tiny-mole [99]

Explanation:

oxidation of Nitrogen in NO2 is +4

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